Answer to Question #334327 in Statistics and Probability for jul

Question #334327

From a finite population consisting of digits 1, 4, 5, and 9, construct the


sampling distribution of the sample mean


a. With a sample size of 2


b. With a sample size of 3

1
Expert's answer
2022-04-28T05:24:49-0400

a. We have population values 1,4,5,9, population size N=4 and sample size n=2.

Mean of population "(\\mu)" = "\\dfrac{1+4+5+9}{4}=4.75"

Variance of population 


"\\sigma^2=\\dfrac{\\Sigma(x_i-\\bar{x})^2}{n}"


"=\\dfrac{14.0625+0.5625+0.0625+18.0625}{4}=8.1875"

"\\sigma=\\sqrt{\\sigma^2}=\\sqrt{8.1875}\\approx2.8614"

The number of possible samples which can be drawn without replacement is "^{N}C_n=^{4}C_2=6."

"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n no & Sample & Sample \\\\\n& & mean\\ (\\bar{x})\n\\\\ \\hline\n 1 & 1,4 & 2.5\\\\\n \\hdashline\n 2 & 1,5 & 3 \\\\\n \\hdashline\n 3 & 1,9 & 5 \\\\\n \\hdashline\n 4 & 4,5 & 4.5 \\\\\n \\hdashline\n 5 & 4,9 & 6.5 \\\\\n \\hdashline\n 6 & 5,9 & 7 \\\\\n \\hdashline\n\\end{array}"




"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n \\bar{X} & f(\\bar{X}) &\\bar{X} f(\\bar{X}) & \\bar{X}^2f(\\bar{X})\n\\\\ \\hline\n 2.5 & 1\/6 & 2.5\/6 & 6.25\/6 \\\\\n \\hdashline\n 3 & 1\/6 & 3\/6 & 9\/6 \\\\\n \\hdashline\n 4.5 & 1\/6 & 4.5\/6 & 20.25\/6 \\\\\n \\hdashline\n 5 & 1\/6 & 5\/6 & 25\/6 \\\\\n \\hdashline\n 6.5 & 1\/6 & 6.5\/6 & 42.25\/6 \\\\\n \\hdashline\n 7 & 1\/6 & 7\/6 & 49\/6 \\\\\n \\hdashline\n\\end{array}"


Mean of sampling distribution 

"\\mu_{\\bar{X}}=E(\\bar{X})=\\sum\\bar{X}_if(\\bar{X}_i)=4.75=\\mu"


The variance of sampling distribution 

"Var(\\bar{X})=\\sigma^2_{\\bar{X}}=\\sum\\bar{X}_i^2f(\\bar{X}_i)-\\big[\\sum\\bar{X}_if(\\bar{X}_i)\\big]^2""=\\dfrac{151.75}{6}-(4.75)^2=\\dfrac{16.375}{6}= \\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})""\\sigma_{\\bar{X}}=\\sqrt{\\dfrac{131}{48}}\\approx1.6520"

b. We have population values 1,4,5,9, population size N=4 and sample size n=3.

Mean of population "(\\mu)" = "\\dfrac{1+4+5+9}{4}=4.75"

Variance of population 


"\\sigma^2=\\dfrac{\\Sigma(x_i-\\bar{x})^2}{n}"


"=\\dfrac{14.0625+0.5625+0.0625+18.0625}{4}=8.1875"

"\\sigma=\\sqrt{\\sigma^2}=\\sqrt{8.1875}\\approx2.8614"

The number of possible samples which can be drawn without replacement is "^{N}C_n=^{4}C_3=4."

"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n no & Sample & Sample \\\\\n& & mean\\ (\\bar{x})\n\\\\ \\hline\n 1 & 1,4,5 & 10\/3\\\\\n \\hdashline\n 2 & 1,4,9 & 14\/3 \\\\\n \\hdashline\n 3 & 1,5,9 & 15\/3 \\\\\n \\hdashline\n 4 & 4,5,9 & 18\/3 \\\\\n \\hdashline\n\\end{array}"




"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n \\bar{X} & f(\\bar{X}) &\\bar{X} f(\\bar{X}) & \\bar{X}^2f(\\bar{X})\n\\\\ \\hline\n 10\/3 & 1\/4 & 10\/12 & 100\/36 \\\\\n \\hdashline\n 14\/3 & 1\/4 & 14\/12 & 196\/36 \\\\\n \\hdashline\n 15\/3 & 1\/4 & 15\/12 & 225\/36 \\\\\n \\hdashline\n 18\/3 & 1\/4 & 18\/12 & 324\/36 \\\\\n \\hdashline\n\\end{array}"


Mean of sampling distribution 

"\\mu_{\\bar{X}}=E(\\bar{X})=\\sum\\bar{X}_if(\\bar{X}_i)=4.75=\\mu"


The variance of sampling distribution 

"Var(\\bar{X})=\\sigma^2_{\\bar{X}}=\\sum\\bar{X}_i^2f(\\bar{X}_i)-\\big[\\sum\\bar{X}_if(\\bar{X}_i)\\big]^2""=\\dfrac{845}{36}-(4.75)^2=\\dfrac{32.75}{36}= \\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})""\\sigma_{\\bar{X}}=\\sqrt{\\dfrac{32.75}{36}}\\approx0.9538"


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